Difference between revisions of "2014 Canadian MO Problems/Problem 1"
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| + | == Problem 1 == | ||
| + | Let <math>a_1,a_2,\dots,a_n</math> be positive real numbers whose product is <math>1</math>. Show that the sum <math>\textstyle\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots(1+a_n)}</math> is greater than or equal to <math>\frac{2^n-1}{2^n}</math>. | ||
| + | == Solution == | ||
<math>\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\=(1-\frac{1}{1+a_1})+(\frac{1}{1+a_1}-\frac{1}{(1+a_1)(1+a_2)})+\cdots+(\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)})\\=1-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\\geq 1-\frac{1}{(2\sqrt{1\cdot a_1)}(2\sqrt{1\cdot a_2)}\cdots (2\sqrt{1\cdot a_n)}}\\=1-\frac{1}{2^n}\\=\frac{2^n-1}{2^n}</math> | <math>\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+\cdots+\frac{a_n}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\=(1-\frac{1}{1+a_1})+(\frac{1}{1+a_1}-\frac{1}{(1+a_1)(1+a_2)})+\cdots+(\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)})\\=1-\frac{1}{(1+a_1)(1+a_2)\cdots (1+a_n)}\\\geq 1-\frac{1}{(2\sqrt{1\cdot a_1)}(2\sqrt{1\cdot a_2)}\cdots (2\sqrt{1\cdot a_n)}}\\=1-\frac{1}{2^n}\\=\frac{2^n-1}{2^n}</math> | ||
Revision as of 10:54, 28 March 2019
Problem 1
Let 
be positive real numbers whose product is 
. Show that the sum 
is greater than or equal to 
.
Solution
